Practice Test 3 solutions - ISyE 3232C Stochastic...

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ISyE 3232C Stochastic Manufacturing and Service Systems Spring 2010 Practice Test 3 (Solution) Closed book, closed notes, no electronics. 1. Let X p be the service time for a perm, and X m be the service time for a manicure. Then, X p exp(2) and X m exp(3). Also, let T S ,T L ,T G be the service time of Sue, Liz and Grace, respectively. Note that Y L = X p ,Y G = X m and Y S = X p . (a) E [min { Y L ,Y G } ] + E [ Y S ] = 1 2 + 3 + 1 2 = 7 10 (b) P ( Y L < Y G ) P ( Y S > Y G ) + P ( Y L > Y G ) P ( Y S > Y L ) = 2 2 + 3 3 2 + 3 + 3 2 + 3 2 2 + 2 = 27 50 (c) P ( Y L > Y G ) P ( Y L > Y S ) = 3 2 + 3 2 2 + 2 = 3 10 2. Let X A be the remaining lifetime of person A , and Y be the interarrival time of kidney arrivals. Then, X A exp( μ A ) and Y exp( λ ). Then, the probability that A obtains a new kidney is same as P ( X A > Y ) = λ λ + μ A . 3. { N ( t ) : t 0 } is a P.P( λ = 30/hour). (a)
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This note was uploaded on 04/29/2011 for the course ISYE 3232 taught by Professor Billings during the Spring '07 term at Georgia Institute of Technology.

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Practice Test 3 solutions - ISyE 3232C Stochastic...

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